Apply inverse definite minimum$-$time $($IDMT$)$ relay with normal inverse characteristic given as

$t=\frac{0\mathrm{.}14}{(\frac{\text{I}}{I_S}{)}^{0\mathrm{.}02}-1}{T}_P$

Where $t=$ tripping time of relay, $T_p=$ time

Network parameters: $G$ : $2$MVA,$10\%$;

$T_1$ : $2$MVA,$\frac{11}{3\mathrm{.}3}KV,7\%$;$,T_2$:$,500$KVA,

$3\mathrm{.}3/0\mathrm{.}415$KV$,6\%$; Lines $1,2,3$: $0\mathrm{.}02\frac{}{p}h$;

Line $5$: $0\mathrm{.}2\frac{}{p}h$; Line $8$: $0\mathrm{.}2\frac{}{p}h$; System MVA

base $=20$MVA; Positive and negative sequence

impedances are assumed equal.

In this network, for a fault at the point $F_1,$ relays $R_1,R_2$ and $R_3$ must not operate in less than $0\mathrm{.}5$ s $.$

Reactance of $G=1p.u$; Reactance of $T_1=0\mathrm{.}7$ p$.$u; Reactance of Line $8=0\mathrm{.}367$ p$.$u; Reactance of Line $5=0\mathrm{.}367$ p$.$u; Reactance of $T_2=2\mathrm{.}4$p$.$u; Reactance of Line $1=2\mathrm{.}323$ p$.$u;

For example, the feeder relay and breaker, which are downstream, should clear a feeder fault before the supply$-$side relay and breaker $($upstream$)$ could trip. Generally, CTI of about $0\mathrm{.}4$ seconds

$2$